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Gekata [30.6K]
3 years ago
6

Assume that a company hires employees on Mondays, Tuesdays, or Wednesdays with equal likelihood.a. If two different employees ar

e randomly​ selected, what is the probability that they were both hired on a Monday​?b. If two different employees are randomly​ selected, what is the probability that they were both hired on the same day of the week​?c. What is the probability that 7 people in the same department were all hired on the same day of the week​?d. Is such an event​ unlikely?
Mathematics
1 answer:
horsena [70]3 years ago
6 0

Answer:

P(A) = \frac{1}{9}

P(B) = \frac{1}{3}

P(C) = \frac{1}{729}

Step-by-step explanation:

We know that:

Only employees are hired during the first 3 days of the week with equal probability.

2 employees are selected at random.

So:

A. The probability that an employee has been hired on a Monday is:

P(M) = \frac{1}{3}.

If we call P(A) the probability that 2 employees have been hired on a Monday, then:

P(A) =P(M\ and\ M)\\\\P(A)=( \frac{1}{3})(\frac{1}{3})\\\\P(A) = \frac{1}{9}

B. We now look for the probability that two selected employees have been hired on the same day of the week.

The probability that both are hired on a Monday, for example, we know is P(A) = \frac{1}{9}. We also know that the probability of being hired on a Monday is equal to the probability of being hired on a Tuesday or on a Wednesday. But if both were hired on the same day, then it could be a Monday, a Tuesday or a Wednesday.

So

P(B) = \frac{1}{9} + \frac{1}{9} + \frac{1}{9}\\\\P(B) = \frac{1}{3}.

C. If the probability that two people have been hired on a specific day of the week is 3(\frac{1}{3}) ^ 2, then the probability that 7 people have been hired on the same day is:

P(C) = (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7 + (\frac{1}{3}) ^ 7\\\\P(C) = \frac{1}{729}

D. The probability is \frac{1}{729}. This number is quite close to zero. Therefore it is an unlikely bastate event.

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